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Topic: Drive by sight (Read 5760 times)

I noticed that there is some change in drive by sight mode. Previously the trains just reserved one tile ahead, now it seem that the driver can see much further ahead, almost up to the next corner or hill. It seems also to affect reservations on other signalling methods, and makes debugging a bit more difficult. So, please what are the current "line-of-sight" rules, and are they in final version?

It appears that the sighting distance is set in simuconf (currently 875meters). This is how far a train can see signals and stuff, meaning that if the train see a signal 800 meters ahead being clear, it will pickup speed.

From quick testing, this is what I found:Straight n-s e-w tracks: sighting distance (SD) is 875 meters.Curve (90 or 45 degre): SD until curve entry but not the curve exit. Curve exit only visible when train is on curve entry tile.Slope: SD until the tile BEFORE the slope, both when slope is pointing upward or downard. While on the slope: SD of the rest of the slope (if more than one tile of slope) but not the tracks at the target level. Maybe there are room for improvement here, since you logically would be able to se an upgoing slope in advance just as well as normal straight tracks? The same would apply to when it is going on the slope down, it would make sence if the train could see the leveled out tracks?Diagonal tracks: Something seems buggy at diagonal tracks: It does sometimes reserve the full SD at ne-sw tracks, but it is not refreshing itself properly, meaning that the SD will appear shorter. at the other diagonal tracks (se-nw) there appears to be no SD.Going under bridges: SD until tile before bridge. When train enter tile AFTER the bridge (not under), SD is 875 meters again.Going into tunnels: No effect on SD

That is odd.. Now the savegame with the testtrack work with no bugs on the diagonal tracks! Maybe I was playing around with signalling too much (made some other testing in the same map) so something went wrong? Anyway, will let you know if I find it again!

However, what about the sighting distance on slope hills, being able to see further?

Do you have any idea why trams have a shorter braking distance than trains? I may well need to adjust their brake force; would you happen to have any figures? I will need to give some thought as to how this interacts with drive by sight.

Incidentally, in relation to the earlier part of this thread, as you may have noticed, I did implement the effect of hills and corners on the sighting distance.

I would just guess that trams are lighter than trains, and have shorter cars = more axles per meter = more brakes 26 m long express train car (Ampz) has 4 axles, 47.5 tons30 m long tram kt8d5 has 8 axles, 38 tfor emergency braking the electromagnetic brakes are long time standard for trams, but not so common on trains.

And one more thing - trams usually have an engine on each axle, and can brake using the engine (recuperate)

Slightly off topic but I think the reason trains are usually heavier than trams of similar capacity would be because they are designed to travel faster. A lot of weight on trains is structural to try and preserve integrity during collisions and other accidents which can occur at high speeds. Since trams are inherently much slower they do not need to be anywhere near as strong since they will usually only be involved in comparatively low speed accidents.

Additionally if trams are narrower than trains (might have to be to fit around cities) then they will have less weight because they are carrying less per meter length.

Maybe the drive by sight maximum speed per rail type should be configurable as a game setting? This would reflect the law like nature it is implemented as in some countries I would think.

Weight and number of axles have little influence on braking deceleration. More mass means more braking force, but also more inertia. On all rail vehicles with wheel brakes, the braking force has been tuned to the mass of the vehicle to give, at maximum braking, a deceleration set by the friction coefficient between steel track and steel wheels. (Goods wagons, the European ones at least, have a switch to tune the brakes to either full or empty wagon.) But when using magnetic rail brakes (those brakes that use an electromagnet to pull a brake shoe against the rails), mass doesn't influence braking force and larger deceleration is possible. Trains seldom use these, but they are more common on trams. In NL some trains have electromagnetic rail brakes, but only use them in emergencies or as parking brake, as they interfere with the safety system (and damage the track if used on points). My local tramway (closed in 1955) used those brakes from shortly after the beginning in 1911 to deal with a 1:12 slope. You won't find many slopes like that on a railway.

Furthermore, trains rarely have brakes with ABS on all axles and the reaction time of train brakes, especially on goods trains, is much longer than on trams, often several seconds.

Slightly off topic: another reason why trams are lighter than trains is that they are designed for lower comfort level.

Thank you all for your replies: I have changed the code so as to exempt trams from the max_speed_drive_by_sight limit. I am still a little unclear on how to calculate the brake force from the data given, however: is anyone able to assist? Or perhaps somebody could give a range of likely brake forces for trams of different eras that I could apply, perhaps after testing performance in the game?

I found on german wikipedia (here) that since 1960 german trams must be capable of 2.73 m/s2 deceleration, so that gives a braking force of F=m×2.73, using newtons and kilogrammes (or kN and tonnes). m is the mass of the vehicle, which must be the maximum mass (i.e., when overcrowded to the limit). That would be the braking force using the emergency brake, which is what we need for the drive by sight speed limit. Of course, when braking for a station stop, the tram would brake less hard, as a sudden 2.73 m/s2 deceleration would certainly lead to injured passengers.

If we have actual braking distances, then instead of assuming the tram follows german law we can calculate its braking force as:F=0.5×m×v2/dwith F force, m mass, v speed, d braking distance. There is some air drag and magnetic friction in this, and those are speed dependent, so this is at best an approximation.